Problem: Simplify and expand the following expression: $ \dfrac{5y + 6}{3y - 3}-\dfrac{4y + 1}{4y + 9} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3y - 3)(4y + 9)$ Multiply the first term by $\dfrac{4y + 9}{4y + 9}$ $ \begin{align*} \dfrac{5y + 6}{3y - 3} \times \dfrac{4y + 9}{4y + 9} & = \dfrac{(5y + 6)(4y + 9)}{(3y - 3)(4y + 9)} \\ & = \dfrac{20y^2 + 69y + 54}{(3y - 3)(4y + 9)}\end{align*} $ Multiply the second term by $\dfrac{3y - 3}{3y - 3}$ $ \begin{align*} \dfrac{4y + 1}{4y + 9} \times \dfrac{3y - 3}{3y - 3} & = \dfrac{(4y + 1)(3y - 3)}{(4y + 9)(3y - 3)} \\ & = \dfrac{12y^2 - 9y - 3}{(4y + 9)(3y - 3)}\end{align*} $ Now we have: $ = \dfrac{20y^2 + 69y + 54}{(3y - 3)(4y + 9)} - \dfrac{12y^2 - 9y - 3}{(4y + 9)(3y - 3)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{20y^2 + 69y + 54 - (12y^2 - 9y - 3)}{(3y - 3)(4y + 9)} $ $ = \dfrac{20y^2 + 69y + 54 - 12y^2 + 9y + 3}{(3y - 3)(4y + 9)} $ $ = \dfrac{8y^2 + 78y + 57}{(3y - 3)(4y + 9)}$ Expand the denominator: $ = \dfrac{8y^2 + 78y + 57}{12y^2 + 15y - 27}$